how to do definite integration when the integrand has a constant term

How to find the value of this integral in matlab ? q should remain as it is in the final answer as I is a function of q

sigma= 1/(1.5e-12)

L=6.4747e3

k=2*pi*1.8331/(390e-12)

D=2.3167e-4

omega sholud me from 379.62 to 389.62

[Walter Roberson]

Symbolic integration?

[Akriti Raj]

Yes, can you please write the matlab code. I am new to matlab and I have been struggling with this problem since long.

[Walter Roberson]

I think I need a higher resolution image of the equation.

Also please confirm that I is a function of the derivative of q as it appears to say I(q') =

[Akriti Raj]

I is a function of q

[Akriti Raj]

[Walter Roberson]

format long g
Q = @(v) sym(v);     %convert to rational

Pi = Q(pi);
sigma = 1/(Q(1.5)*10^-12)

sigma = 

L = Q(64747)/10

L = 

k = 2 * Pi* Q(18331)/10^4/(Q(390)*10^-12)

k = 

D = Q(23167)*10^-8

D = 

omega_min =  Q(37962)/100

omega_min = 

omega_max = Q(38962)/100

omega_max = 

syms q
syms omega

inner = exp((omega/sigma)^2/2) * sinc(L*q^2/k + D*omega*L/2).^2

inner = 

I = 1/(sigma * sqrt(2*Pi)) * int(inner, omega, omega_min, omega_max)

I = 

simplify(I)

ans = 

inner_approx = taylor(inner, omega, (omega_min+omega_max)/2, 'order', 4)

inner_approx = 

I_approx = 1/(sigma * sqrt(2*Pi)) * int(inner_approx, omega, omega_min, omega_max)

I_approx = 

simplify(I_approx)

ans = 

fplot(I, [-5 5]); title('exact')

double(subs(I,q,[-5 -1 0 1 5]))

ans = 1×5

3.49318454188821e-18 3.49318582811878e-18 3.4931858817153e-18 3.49318582811878e-18 3.49318454188821e-18

vpa(subs(I,q,[-5 -1 0 1 5]*10^9))

ans = 

fplot(I_approx, [-5 5]); title('approximated')

double(subs(I_approx,q,[-5 -1 0 1 5]))

ans = 1×5

-3.2139136293569e-16 -3.21388912778066e-16 -3.21388810680452e-16 -3.21388912778066e-16 -3.2139136293569e-16

vpa(subs(I_approx,q,[-5 -1 0 1 5]*10^9))

ans = 

[Walter Roberson]

Notice that up in the 10^9 area, that the integral could not be calculated with the default integration options.

Note that the taylor approximation is probably pretty wrong by the time of 10^9

What is the expected range of q ?

[David Goodmanson]

Hi Akriti,

since you refer to this as a definite integral, what are the limits of integration?

[Walter Roberson]

@David Goodmanson

"omega sholud me from 379.62 to 389.62"

[David Goodmanson]

thanks Walter I missed that

[Akriti Raj]

@walter q should be between -1 to 1